Could you please help me with this integral? $$\int_0^1\frac{x^{42}}{\sqrt{x^4-x^2+1}} \operatorname d \!x$$
Update: user153012 posted a result given by a computer that contains scary Appel function, and Cleo gave much simpler closed forms for powers $n=42,\,43$. I am looking for a way to prove those forms. I also would like to find a more general result that would work for arbitrary integer powers, not just $42$.
On
$$\int_0^1\frac{x^{42}}{\sqrt{x^4-x^2+1}}dx=\frac{250\,351\,656\,060\,403}{1\,955\,894\,551\,246\,350}\\-\frac{25\,556\,904\,389\,521}{391\,178\,910\,249\,270}{\bf K}\left(\frac{\sqrt3}2\right)+\frac{29\,595\,166\,842\,073}{977\,947\,275\,623\,175}{\bf E}\left(\frac{\sqrt3}2\right),$$
where ${\bf K}(x)$ and ${\bf E}(x)$ are complete elliptic integrals of the $1^{st}$ and $2^{nd}$ kind.
You might be also interested to know that $$\int_0^1\frac{x^{43}}{\sqrt{x^4-x^2+1}}dx=\frac{10\,495\,168\,793\,593}{86\,586\,540\,687\,360}-\frac{98\,084\,055\,671}{1\,099\,511\,627\,776}\ln3.$$
On
Mathematica and Maple could solve this integral in term of elliptic integrals. Cleo's answer is a simplification of them.
If you prefer there is an other closed form in term of Appell $F_1$ function.
$$I(a,n)=\int_0^1\frac{x^{a}}{\sqrt[n]{x^4-x^2+1}}\,dx = \frac{1}{1+a} F_1\left(\frac{1+a}{2},\frac{1}{n},\frac{1}{n},\frac{3+a}{2},(-1)^{1/3},-(-1)^{2/3}\right),$$ where $\Re(a)>-1.$
Your case is $I(42,2)$.
Odd case: The change of variables $x^2=t$ transforms the integral into $$\mathcal{I}_{2n+1}=\int_0^1\frac{x^{2n+1}dx}{\sqrt{x^4-x^2+1}}=\frac12\int_0^1\frac{t^ndt}{\sqrt{t^2-t+1}}$$ Further change of variables $t=\frac12+\frac{\sqrt3}{4}\left(s-\frac1s\right)$ allows to write $t^2-t+1=\frac3{16}\left(s+\frac1s\right)^2$ and therefore gives an integral of a simple rational function of $s$: $$\mathcal{I}_{2n+1}=\frac12\int_{1/\sqrt3}^{\sqrt3}\left[\frac12+\frac{\sqrt3}{4}\left(s-\frac1s\right)\right]^n\frac{ds}{s}.$$
Even case: To demystify the result of Cleo, let us introduce $$\mathcal{K}_n=\mathcal{I}_{2n}=\int_0^1\frac{x^{2n}dx}{\sqrt{x^4-x^2+1}}=\frac12\int_0^1\frac{t^{n-\frac12}dt}{\sqrt{t^2-t+1}}.$$ Note that $$\mathcal{K}_{n+1}-\frac12\mathcal{K}_n=\frac12\int_0^1 t^{n-\frac12}d\left(\sqrt{t^2-t+1}\,\right)=\frac12-\left(n-\frac12\right)\left(\mathcal{K}_{n+1}-\mathcal{K}_{n}+\mathcal{K}_{n-1}\right),$$ where the second equality is obtained by integration by parts. This gives a recursion relation $$\left(n+\frac12\right)\mathcal{K}_{n+1}=n\mathcal{K}_{n}-\left(n-\frac12\right)\mathcal{K}_{n-1}+\frac12,\qquad n\geq1.$$ It now suffices to show that \begin{align*} \mathcal{K}_0&=\int_0^1\frac{dx}{\sqrt{x^4-x^2+1}}=\frac12\mathbf{K}\left(\frac{\sqrt3}{2}\right),\\ \mathcal{K}_1&=\int_0^1\frac{x^2dx}{\sqrt{x^4-x^2+1}}= \frac12\mathbf{K}\left(\frac{\sqrt3}{2}\right)-\mathbf{E}\left(\frac{\sqrt3}{2}\right)+\frac12. \end{align*}